Talk:Schwarzschild radius

Incorrect derivation?

http://scienceworld.wolfram.com/physics/SchwarzschildRadius.html claims the derivation here is incorrect. that is it is incorrect to set the escape velocity to c the speed of light to solve for r. On consideration, it is OBVIOUS that there is something wrong with an equation (for a finite mass) that claims v = c, isn't there? (since this is impossible.) —Preceding unsigned comment added by 69.40.245.3 (talk) 22:12, 23 August 2008 (UTC)

See the note I just added above (before I noticed this comment!) I believe wolfram is wrong.(no surprise. I used SMP years ago. Maxima, now...) If I drop a rock from infinity (give it an infinitesimal shove to get it going...) then it will accelerate towards the mass (M, at the origin), according to newton's laws, to arbitrary accuracy. Relativistically its mass will increase (relative to a frame at the origin), but that mass cancels out (little m in my note above -- replace m by m(t) or m(x) -- parameterized how you like - it still cancels out.) It's more than a coincidence that the argument works. The exact point where that mass (relativistically) becomes infinite, is where the argument may fail - but that's because that is what it means to have an event horizon.

Steve —Preceding unsigned comment added by 67.90.34.130 (talk) 18:39, 24 March 2011 (UTC)

Needs square-root?

Shouldn't the radius = square root ( 2GM/c^2) ???

No, see [1], [2], [3], etc. -- Tim Starling 04:25, Nov 18, 2004 (UTC)

Or, more simply, using dimensional analysis: $r\,$ is in meters ($m\,$), so $2Gm/c^{2}\,$ must be in meters also. Let's see:

$G\to N\cdot m^{2}/kg^{2}\,$, where $N\,$ is force = mass * acceleration = $kg\cdot m/s^{2}\,$, so $G\,$ is in units of $(kg\cdot m/s^{2})\cdot (m^{2}/kg^{2})\,$.

$m/c^{2}\,$ is in units of $kg/(m/s)^{2}\to kg\cdot s^{2}/m^{2}$.

So when combined $Gm/c^{2}\,$ is in units of $(kg\cdot m/s^{2})\cdot (m^{2}/kg^{2})\cdot (kg\cdot s^{2}/m^{2})$.

You can see directly that units of $kg\,$ and $s\,$ are equal in numerator and denominator, leaving $m^{3}/m^{2}\to m\,$ (meters).

QED. Slowmover 20:23, 22 March 2006 (UTC)

Meaning for non-black holes?

What is the physical meaning of an object's Schwarzschild radius when it is bigger than the radius? -- Myria 05:35, 29 December 2005 (UTC)

It means that this object is not a black hole. For an object to be a black hole, its mass needs to be concentrated in a radius less than its Schwarzschild radius. The same holds also for part of the object, so in theory it's possible that the whole object is not a black hole, but its inner part is a black hole. This subject is important to define the moment when a collapsing star becomes a black hole. This transition happens when its mass is concentrated in a radius that is equal to its Schwarzschild radius.

I'm not exactly where to put it, but when reading this article, it took me a while to figure out that a plain-English definition of the Schwartzchild radius makes more sense. The Schwartzchild radius is the radius for a given mass where, if that mass could be compressed to fit within that radius, no force could stop it from continuing to collapse into a singularity. Wouldn't it be nice to put such a definition in the article? Problem is that I'm not even remotely a physicist so I don't want to make the change myself.--MikeGinny 04:57, 30 June 2006 (UTC)

Giving this some thought. -- Slowmover 19:34, 30 June 2006 (UTC)

"An object smaller than its Schwarzschild radius is called a black hole" ... ok, physics newb here, but does this mean that the universe was technically a black hole in the early stages of the big bang, or did the SR scale with the expansion of the universe? Snaxalotl (talk) 18:32, 7 June 2009 (UTC)snaxalotl

I think that one useful way to look at it is this: If you take an object with a particular mass, it has a certain amount of gravity, all of which has to exit through its surface. If you then squish the object into a smaller volume, the intensity of the gravitational field at the surface increases, and when you increase this surface gravity, you increase the escape velocity, the speed at which somethng has to be thrown with at the surface to have a chance of completely escaping. For a given mass M, the r=2M radius tells us how compact the material would have to be in order for this surface escape velocity to reach the critical situation of being equal to the nominal speed of light. John Michell did the calculations for this back in the Eighteenth Century, but when Isaac Newton's original ideas about light took a hammering, the idea became unfashionable and dropped out of sight until Karl Schwarzchild brought it back, more unambigiuously, in the context of Einstein's general theory of relativity, in the early Twentieth Century. ErkDemon 15:59, 29 December 2006 (UTC)

That is no answer to Snaxalotl's question: Was the universe a black hole in its early stages? 145.97.222.84 (talk) 17:39, 15 January 2012 (UTC)

Gravitational versus Schwarzchild radius

I added a link to the hyperphysics sight because it distinguishes between a gravitational radius and a Scwarzschild radius. The distinction is an important one even though it is quite common to confuse the two.Lucretius 07:53, 15 January 2006 (UTC)

I subsequently removed the link because 'gravitaional radius' and 'Schwarzschild radius' are considered synonymous by the great majority of the scientific community.Lucretius 23:50, 15 January 2006 (UTC)

Low density black hole

below are the calculations for the mass and radius of a black hole with the density of water. The resulting mass is 135 million times the mass of the Sun. The radius is 400 million kilometres.

Calculations

That is the density of an object that perfectly fits its Schwarzschild radius. Pity that, once the object fits the object it will collapse into a singularity of infinite density. From the article: The Schwarzschild radius is the radius for a given mass where, if that mass could be compressed to fit within that radius, no known force or degeneracy pressure could stop it from continuing to collapse into a gravitational singularity -- which according to General Relativity means it will reach infinite density.

I'm not sure about what you wanted to prove. --badpazzword (talk) 08:25, 29 March 2008 (UTC)

I think the point here is only that the density claimed is the average density for everything inside the Schwarzschild radius. A nerd fact for people who like trivia. —Preceding unsigned comment added by Snaxalotl (talk • contribs) 18:25, 7 June 2009 (UTC)

Great Job on the Article

I just wanted to let the editors of this article know this article's far better than it was before; I didn't have a good idea of the Schwarzchild radius earlier, but it's clear now. This seems to be applicable to a lot of the physics articles in particular. Just wanted to point it out- thanks a lot for your contributions! Robinson0120 01:47, 17 January 2007 (UTC)

Removed incorrect claim

I removed this sentence:

The Schwarzschild radius of a sphere with a uniform density equal to the critical density is equal to the radius of the visible universe.

because it's not true--the Schwarzschild radius in this case is about 250 Gly, and the radius of the visible universe is about 50 Gly. There's certainly a meaningful relationship here, but it's not quite so simple. (And I'm not sure exactly what it is.) -- BenRG 16:11, 26 March 2007 (UTC)

Underlying symmetry?

From the article: "Note that although the result is correct, general relativity must be used to properly derive the Schwarzschild radius. Some consider it to be only a coincidence that Newtonian physics produces the same result, yet this may be an indication of a deeper underlying symmetry in nature."

The first sentence and the first half of the second sentence are my own contribution from a while back. However, someone seems to have mangled the message by using the much weaker phrase, "Some consider it..." and the highly questionable, "yet this may be an indication of a deeper underlying symmetry in nature." I won't blindly revert it to, "It is only a coincidence that Newtonian physics produces the same result," but I'm quite skeptical about this "underlying symmetry" they're referring to. Newtonian physics is an approximation, plain and simple. Saying there's a symmetry between general relativity and Newtonian physics is like saying there is a symmetry between cars and engines. Cars encompass engines-- they're not on equal footing. You can compare different models of cars, just as you can compare general relativity to less reputable theories. If someone knows what this underlying symmetry is, I'd love to hear about it. Otherwise, I'll go ahead and revert that sentence to its much clearer form.--134.173.200.14 20:13, 3 April 2007 (UTC)

Rotating objects

Is it possibly so that if an objects rotates fast you could puch some more mass within the S-radius without the object collapsing? (As Schwarzschild is, as I understand, a solution for the non-rotating case.)If so the first paragraph is not entirely correct. -- Agge1000 (talk) 18:44, 16 November 2007 (UTC)

how about "the size of a finger nail."? finger tip? —Preceding unsigned comment added by 69.40.245.3 (talk) 22:14, 23 August 2008 (UTC)

9 mm Walnut?

Quote from the article:

while the Earth's is only about 9 mm, the size of a walnut.

Most walnuts I have known are quite a bit larger than 9 mm. Perhaps peanut would be more appropriate. --66.92.218.124 (talk) 19:28, 22 May 2008 (UTC)

A peanut would be the right size, but the wrong shape. A hazelnut would be a better nut comparison though these can still be significantly larger than 9mm. However if we are considering legumes, I would suggest that chickpeas display more of the correct characteristics. —Preceding unsigned comment added by 193.129.242.5 (talk) 09:54, 13 June 2008 (UTC)

A metric for gravitational mass

	$r_{s}$
Sun	2953.2500765 m
Mercury	0.49028 mm
Venus	7.22909 mm
Earth	8.870056078 mm
Mars	0.95305 mm
Jupiter	2.81915558 m
Saturn	0.84408275 m
Uranus	0.1289325 m
Neptune	0.1521333 m

The International System of Units (SI) has a unique unit for inertial mass (the kilogram) and a unique unit for energy (the Joule), but it does not have a unique unit for gravitational mass. Therefore, gravitational masses are commonly listed in terms of inertial mass equivalents: for example, the mass of the earth as 5974.2 yottagrams. However, this tradition is unfortunate, and even bizarre, when one considers that there is no known way to measure the inertial mass of a planet sized object.

The masses of planets and stars are currently estimated by first measuring their gravity, and then using G in Newton's theory of gravity to estimate the inertial mass. In fact, the inertial masses of the sun and planets have never been directly measured. The values listed in the literature are just estimates using Newton's theory, and there is no known way to prove that these estimates are correct.

See the following article:

"http://curious.astro.cornell.edu/question.php?number=452"

As an alternative, I would propose using the “standard gravitational parameter” to directly calculate the Schwarzschild radius. The fact that c is defined exactly allows the Schwarzschild radius of a gravitating object to be determined to the same accuracy to which one knows the gravity of the object (See the table to the right: the “standard gravitational parameters” of the sun and earth are known to over 9 digits). And using the “standard gravitational parameter” to directly calculate the Schwarzschild radius removes the unnecessary step of estimating the Newtonian inertial mass. Furthermore, the Schwarzschild radius can be indirectly measured by measuring time dilation near the object. So the Schwarzschild radius has the advantage of being both accurate and measurable, whereas the inertial mass is neither measurable nor accurately determinable.

Current alternatives for listing precise gravitational masses are to list mass ratios as is done in this NASA web site, or to list large masses in terms of their standard gravitational parameter. 70.176.112.179 (talk) 04:54, 19 April 2009 (UTC)

Supermassive Black Hole at the center of the Milky Way

Images toward the center of the Milky Way as seen by Hubble show that there is so much galactic matter, both stars and dust, between us and the center that it is impossible to see what is at the center. No black hole has ever been seen at the center of our galaxy, nor any galaxy. The luminosity at the center of galaxies is so great that detail cannot be discerned. That black holes exist at the center of galaxies is pure speculation. My Flatley (talk) 18:27, 1 October 2009 (UTC)

If you look at Supermassive black hole they discuss the evidence for such objects there, stating that there is a consensus amongst astronomers that such things do exist. Read about the evidence, and take your discussion to their talk page if not convinced.Puzl bustr (talk) 15:12, 6 December 2009 (UTC)

Since 10/1/09, astrophysicits and astronomers have confirmed that tere is a supermassive black hole at the center of the Milky Way Galaxy and have predicted that there is a supermassive black hole at the center of every galaxy. - Brad Watson, Miami 71.196.121.70 (talk) 19:51, 11 July 2011 (UTC)

M-sigma relation

I noticed someone questioned the relationship between core black hole mass and the mass of the galaxy - and this is only an average relationship with much scatter, as evidenced by their example comparing Andromeda and Milky Way. I found that there is a much tighter relationship and an article M-sigma relation discussing it, and the previous work in this area, so added a ref and the wikilink. Puzl bustr (talk) 15:12, 6 December 2009 (UTC)

Schwarzschild radius for the Sun

I found this interesting quote and would like to start a section in the article for similar size comparisons: "The Schwarzschild radius for the Sun is about two miles, 1/200,000th of its current width; for the Earth to become a black hole, it would have to be squeezed into a ball with a radius of one centimetre." --Are we living in a designer universe? --Thorwald (talk) 03:24, 22 November 2010 (UTC)

A singularity is not the same as a black hole

I made a few edits in the intro when I noticed that TWO distinct definitions were given. I ultimately saw that these two definitions were intricately muddled throughout the article.

Unless I have some big misunderstandings, a singularity is not the same as a black hole.

Just because the Schwarzschild radius (an event horizon-like-thing) is at the surface doesn't mean the ability to resist gravitational pressure is all lost and an automatic compression to a singularity ensues. A "singularity" is defined to have a radius of ZERO, black holes generally don't have a surface radius of zero, they just have a lot of mass and are small enough that the Schwarzschild radius is above the surface.

In so many places in the article, the SR is referred to as a function of MASS ONLY, which implies it is NOT related to the ability or inability of the matter to resist the gravity pressure, yet in many places the ability to resist gravity pressure and singularities are muddled in with it all.

So, what's the answer? It's a pretty serious problem and, if you'll forgive me, I'm raising the EXTREMELY DUBIOUS flag for the whole article because of it.

I can't fix it. I don't have enough detailed background, nor enough time, nor enough inclination for that matter. But you people who watch this article CAN fix it, and you need to get on it! It's incredibly bad!

Okay, if I'm out-to-lunch it's no big deal, just explain to me why. For that matter, explain it in the article itself. It is unclear in the article just how the ability to resist gravitational pressure is related to the SR being at the surface, etc. If they are indeed related, please try to make that clear in the article. I'm a really smart guy, and it's unclear to me.  :-).

All the best to you all,

Dave (today on IP 108.7.171.191 (talk) 05:31, 27 May 2011 (UTC))

I looked a little bit up about singularities and is seems that in physics it doesn't necessarily mean surface radius is zero. Instead, it has a number of different meanings, depending on context, related to the center of a black hole. Still though, the intermixing of Rsurface = SR with inability to continue to withstand gravity remains dubious or inadequately explained (prior to my quick-fix edits described below). If "singularity" (and/or the other stuff) is brought back, it should be brought back along with more specific explanation about what "singularity" means in this case. Also more explanation about why a nonsensical thing like "(Rsurface = SR) = collapse" is true if indeed it is true. And, since such a thing is an extraordinary claim, it would need extraordinary and extraordinarily reliable references! :-)

Dave (today on IP 108.7.171.191 (talk) 07:24, 27 May 2011 (UTC)

Taking a shot at fixing it

On somewhat closer inspection, it looks like the only remaining of these muddlements are a sentence remaining in the intro and a bunch woven into the HISTORY section. I am going to remove the History section because that is the quickest and easiest way to correct the problem. I know it's a big hairy deal to delete a section, but if you want to put it back, you are certainly welcome, just please do your best to correct the problem as you do (by removing the muddlements or adding explanation why they aren't muddlements)

Yours,

Dave (today on IP 108.7.171.191 (talk) 05:49, 27 May 2011 (UTC))

Wikipedia bugs in evidence here

I made those changes I described above. But, they took a long time to show up in the article, AND they haven't yet appeared in the edit history page. It's been 5 or 6 days so far. There were other strangenesses while I was doing the edits too (too much to go in to). So if you want to see the edits, click on "My" IP address above, you'll see they were actually made, they just don't show up in the edit history for some reason. All other articles I've played with since then are behaving normally. I am an experienced editor by the way, so I do know when WP is behaving strangely.

Dave (today on another IP: 108.7.162.218 (talk) 18:11, 2 June 2011 (UTC))

The Sun has a Schwarzschild radius of app. 3 km (1.86 miles)

the Sun has a Schwarzschild radius of approximately 3.0 km (1.86 miles). - Brad Watson, Miami 71.196.121.70 (talk) 19:44, 11 July 2011 (UTC)

The top image on the page should be a diagram showing at least one radius

The top image on the page should be a diagram showing at least one radius. The Schwarzschild radius might be a good one to include. I don't know what the diagram shown there is but it doesn't seem to have anything to do with gravity or spacetime or light paths. Michael McGinnis (talk) 06:33, 22 July 2011 (UTC)

The Schwarzschild radius in gravitational time dilation

The formula given was

${\frac {t_{r}}{t}}={\sqrt {1-{\frac {r_{s}}{r}}}}={\frac {1}{2}}$

I removed the 1/2 as it clearly doesn't make sense here, (otherwise tr = 1/2 t regardless of the size of the mass) and doesn't correspond to the formula given in the reference article (http://en.wikipedia.org/wiki/Gravitational_time_dilation) where it says

$t_{0}=t_{f}{\sqrt {1-{\frac {2GM}{rc^{2}}}}}=t_{f}{\sqrt {1-{\frac {r_{0}}{r}}}}$

Neill Jones 87.113.10.142 (talk) 15:09, 16 February 2012 (UTC)

Newtonian derivation

A historical note -- as I recall, it's pointed out in Halliday and Resnick that the equation r=2Gm/c^2 was first derived in the 1700s sometime. I suspect it's given as a problem. Easy enough to look up, if anyone cares.

I derived it myself, in high school, by asking what can be said about an object with escape velocity =c - which is in fact the definition of the event horizon of a black hole.

If you want to reproduce it, remember

f=m a =m d^2x/dt^2 = m dv/dt = m dv/dx dx/dt =   m v dv/dx = GmM/x^2, 

rearrange, and integrate. I used to give this as an extra credit problem when I was a physics TA.

Steve Horne (stephen.f.horne@gmail.com) —Preceding unsigned comment added by 67.90.34.130 (talk) 16:22, 24 March 2011 (UTC)

You are assuming Newtonian physics. In which case, $v_{e}=c$ only gives the boundary from where light cannot escape to infinity without additional forces. Light can still go outside such a boundary. Newtonian physics doesn't have event horizons. You need relativity to derive the Schwarzschild radius in a way that makes sense. 217.121.224.124 (talk) 00:08, 1 April 2012 (UTC)

"Shield of blackness"

"Schwarzschild" translates literally as "black shield" or liberally as "shield of blackness". To a German newcomer to astrophysics this would be an obvious name for the phenomenon, before being told that this is the name of the discoverer/hypothesizer. Not unlikely enticed by his own name, the astonomer Karl Schwarzschild thought harder than others of the possibility of a capsule or shield of blackness around a very dense mass. Half a century had to pass after his death until evidence appeared that this was not just a pipedream. Puddington (talk) 22:57, 17 June 2012 (UTC)

Error in the Milky Way's diameter

The milky way's diameter as mentioned in the parameter table is incorrect. It is mentioned at ~0.2ly whereas wikipedia on the 'Milky way' page states it as 100-120kly. Please make the necessary change. — Preceding unsigned comment added by 115.242.197.202 (talk) 16:27, 9 December 2012 (UTC)

The number quoted in the table is not the Milky Way's diameter, it's its Schwarzschild radius. I didn't check if the number given is correct but it should certainly be much smaller than the diameter. If it was the same we wouldn't be here :-) — HHHIPPO 21:41, 9 December 2012 (UTC)

Table of Schwarzchild radii

I think the table would be a lot more helpful if instead of showing only the Schwarzschild radius, it also showed the actual radius of the objects. If both of these are next to each other, the reader can better appreciate how far those objects are from black holes. Bande-Ali (talk) 07:15, 19 December 2012 (UTC)